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Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread $\noindent Yeah, we shouldn't include $\pi$ and $\frac{5\pi}{4}$, as they are not in the domain of our function.$

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread $\noindent Note that $y^\prime = -\tan x + \sec^2 x -1$. Hence $y^\prime =-\tan x + \tan^2 x$, as $\sec^2 x -1 = \tan^2 x$. Now it is easy to solve $y^\prime =0$. Just make sure to only include solutions $x$ such that $\cos x>0$...

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread $\noindent For (c), if we differentiate the function, we find using parts (a) and (b) that $y^\prime = -1-(-1)=0$, provided that $\cos x >0$ (i.e. $x$ is in the first or fourth quadrants). So each piece of the graph has 0...

But HSC 3U/4U maths never seems to be referred to as 'advanced'. I think the reason for this is that they have names like 'Maths Ext. 1/2', which make them easily identifiable, whereas 2U maths is officially called Mathematics, which makes it ambiguous as to what level of maths it is. So 2U is...

$\noindent What do you mean by that? Do you mean the number $\log _ 2 3$ (we shouldn't be saying `to the power of' 3)? Generally for these Q's, the approach is a proof by contradiction.$ $\noindent Assuming you meant this, assume by way of contradiction that $\log_2 3 =\frac{p}{q}$ for some...

Why is NSW HSC 2U maths often referred to as Advanced Maths? It's not that 'advanced'. Is it simply to distinguish it from HSC General Maths?

Re: HSC 2016 3U Marathon $\noindent Since knowledge of the golden ratio's definition isn't expected in the HSC, I'll provide it here: it's the positive root of the quadratic equation $x^2-x-1=0$.$

$\noindent Wait, why is it $x$? Did you mean $(x-1)$? Because $\binom{x}{x}=1$.$

Yeah, no-one yet knows how to prove or disprove it. This is why it is an unsolved problem. :) And the Riemann Hypothesis one is that all the non-trivial zeros of this special function called the Riemann-Zeta function have real part ½. (The input values of the Riemann Hypothesis are complex...

$\noindent Yes. This is because the graph of $y = \cosh x$ fails the Horizontal Line Test.$

And 2016 has too many factors to do by hand without it getting very tedious. (There could be a more elegant way though.)

And 2016 has too many factors to do by hand without it getting very tedious.

For anyone who might want to try this using 2016's factors, here are its factors: 1, 2, 3, 4, 6, 7, 8, 9, 12, 14, 16, 18, 21, 24, 28, 32, 36, 42, 48, 56, 63, 72, 84, 96, 112, 126, 144, 168, 224, 252, 288, 336, 504, 672, 1008, 2016 .

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread What are the questions?

Well better late than never (there actually were prerequisites in the past, but they were removed).

Re: HSC 2016 3U Marathon Your method wasn't actually quite correct. This is because the events of pairs having different birthdays are not independent events, so we can't just multiply the probabilities like that. If we could just multiply them like that, we'd always have a non-zero...

I think there were many reasons. One reason can be found by reading the first few paragraphs of the article in this post: http://community.boredofstudies.org/586/general-university-discussion/347762/universities-blame-maths-crisis-among-high-school-students.html#post7104785

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread $\noindent Let the two roots be $\alpha, \,3\alpha$. By sum of roots, we have $\alpha+3\alpha = 4\alpha = -\frac{b}{a}\Rightarrow \alpha = -\frac{b}{4a}$. Now, by product of roots, $\alpha \cdot 3\alpha = 3\alpha^2 = \frac{c}{a}...

Perhaps, but I'm guessing it's best not to burst out laughing at them, just keep the thought to yourself I suppose.

The answer is (B), I believe. For these three-point median regression lines, we have to divide the data into three 'clumps', consider a line connecting the medians of the outer two clumps, and then move this one-third of the way towards the median of the inner clump. See here for more detail...